JEE Mains · Maths · STD 11 - 10.1 circle and system of circle
Let \(\mathrm{C}\) be a circle with radius \(\sqrt{10}\) units and centre at the origin. Let the line \(x+y=2\) intersects the circle \(\mathrm{C}\) at the points \(\mathrm{P}\) and \(\mathrm{Q}\). Let \(\mathrm{MN}\) be a chord of \(C\) of length \(2\) unit and slope \(-1\) . Then, a distance (in units) between the chord \(PQ\) and the chord \(MN\) is :
- A \(2-\sqrt{3}\)
- B \(3-\sqrt{2}\)
- C \(\sqrt{2}-1\)
- D \(\sqrt{2}+1\)
Answer & Solution
Correct Answer
(B) \(3-\sqrt{2}\)
Step-by-step Solution
Detailed explanation
\( \mathrm{C}: \mathrm{x}^2+\mathrm{y}^2=10 \) \( \mathrm{AN}=\frac{\mathrm{MN}}{2}=1 \) \( \therefore \mathrm{In} \Delta \mathrm{OAN} \rightarrow(\mathrm{ON})^2=(\mathrm{OA})^2+(\mathrm{AN})^2 \) \( 10=(\mathrm{OA})^2+1 \rightarrow \mathrm{OA}=3\) Perpendicular distance of…
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